[Math] Two questions about finiteness of ideal classes in abstract number rings

Question

Let us say that an abstract number ring is an integral domain $R$ which is not a field, and which has the "finite norms" property: for any nonzero ideal $I$ of $R$, the quotient $R/I$ is finite.

(I have taken to calling such rings abstract number rings and have some vague ambitions of extending the usual algebraic number theory to this class of rings. Note that they include the two basic rings $\mathbb{Z}$ and $\mathbb{F}_p[t]$ and are closed under: localization, passage to an overring — i.e., a ring intermediate between $R$ and its field of ractions — completion, and taking integral closure in a finite degree extension of the fraction field. In order to answer my questions affirmatively one would have to know something about abstract number rings which are not obtained from the two basic rings via any of the above processes — if any!)

Note that such a ring is necessarily Noetherian of dimension one, so it is a Dedekind domain iff it is normal, and in any case its integral closure is a Dedekind abstract number ring.

Question 1: Does there exist an integrally closed abstract number ring with infinite Picard (= ideal class, here) group?

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Question 2: Let $R$ be a not-necessarily integrally closed abstract number ring with integral closure $\tilde{R}$. Suppose that the ideal class group of $\tilde{R}$ is finite. Consider the ideal class monoid $\operatorname{ICM}(R)$ of $R$, i.e., the quotient of the monoid of nonzero ideals of $R$ by the submonoid of principal ideals. (Note that the group of units of $\operatorname{ICM}(R)$ is precisely the Picard group, but if $R$ is not integrally closed it will necessarily have non-invertible ideals so that $\operatorname{Pic}(R)$ will not be all of $\operatorname{ICM}(R)$.) Can it be that $\operatorname{ICM}(R)$ is infinite?

Answer 1

To answer Question 1: Yes, there do exist integrally closed abstract number rings with infinite class group.

By factorization of ideals, for $R$ to be an abstract number ring it is enough that it is a Dedekind domain with finite residue field $R/\mathfrak{p}$ at each prime $\mathfrak{p}$. Theorem B of the paper mentioned by Hagen Knaf in his answer actually gives what you ask for (R. C. HEITMANN, PID’S WITH SPECIFIED RESIDUE FIELDS, Duke Math. J. Volume 41, Number 3 (1974), 565-582).

Theorem B: Let G be a countable abelian torsion group. Then there is a countable Dedekind domain of characteristic 0 whose class group is G, and whose residue fields are those of the integers (i.e. one copy of $\mathbb{Z}/p\mathbb{Z}$ for each prime $p$).

As such rings have finite residue fields, this gives an integrally closed abstract number ring with class group any countable torsion group you like. We can do much better than this though. After thinking about your question for a bit, I see how we can construct the following, so that all countable abelian groups occur as the class group of such rings.

Let G be a countable abelian group. Then, there is a Dedekind domain $R$ with finite residue fields such that $\mathbb{Z}[X]\subseteq R\subseteq\mathbb{Q}(X)$ and ${\rm Cl}(R)\cong G$.

I see some surprise mentioned in the comments below that it is enough to look at over-rings of $\mathbb{Z}[X]$ to find Dedekind domains with any countable class group. In fact, over-rings of $\mathbb{Z}[X]$ are very general in terms of prime ideal factorization, and can show the following. I'll use ${\rm Id}(R)$ for the group of fractional ideals of $R$ and $R_{\mathfrak{p}}$ for the localization at a prime $\mathfrak{p}$, with $\bar R_{\mathfrak{p}}$ representing its completion (which is a compact discrete valuation ring (DVR) in this case).

Let $R$ be a characteristic zero Dedekind domain with finite residue fields. Then, there is a Dedekind domain $R^\prime$ with $\mathbb{Z}[X]\subseteq R^\prime\subseteq\mathbb{Q}(X)$ and a bijection $\pi\colon {\rm Id}(R)\to{\rm Id}(R^\prime)$ satisfying

1. $\pi(\mathfrak{ab})=\pi(\mathfrak{a})\pi(\mathfrak{b})$.
2. $\pi(\mathfrak{a})$ is prime if and only if $\mathfrak{a}$ is.
3. $\pi(\mathfrak{a})$ is principal if and only if $\mathfrak{a}$ is.
4. If $\mathfrak{p}\subseteq R$ is a nonzero prime then $\bar R_{\mathfrak{p}}\cong\bar R^\prime_{\pi(\mathfrak{p})}$.

In particular, the class groups are isomorphic, ${\rm Cl}(R)\cong{\rm Cl}(R^\prime)$.

The idea is that we can construct Dedekind domains in a field $k$ by first choosing a set $\{v_i\colon i\in I\}$ of discrete valuations on $k$ and, letting $k_v=\{x\in k\colon v(x)\ge0\}$ denote the valuation rings, we can take $R=\bigcap_ik_{v_i}$. Under some reasonably mild conditions, this will be a Dedekind domain with the valuations $v_i$ corresponding precisely to the $\mathfrak{p}$-adic valuations, for prime ideals $\mathfrak{p}$ of $R$. In this way, we can be quite flexible about constructing Dedekind domains with specified prime ideals (and, with a bit of work, specified principal ideals and class group). Constructing discrete valuations $v$ on $k=\mathbb{Q}(X)$ is particularly easy. Given a compact DVR $R$ of characteristic 0 and field of fractions $E$, every extension $\theta\colon k\to E$ gives us a valuation $v(f)=u(f(X))$ where $u$ is the valuation in $E$. To construct such an embedding only requires choosing $x\in E$ which is not algebraic over $\mathbb{Q}$ and, if we want the localization $k_v$ to have completion isomorphic to $R$, then we just need $\mathbb{Q}(x)$ to be dense in $E$. There's plenty of freedom to choose $x\in R$ like this. In fact, there's uncountably many $x$, as they form a co-meagre subset of $R$. So, we have many many valuations on $\mathbb{Q}(X)$ corresponding to any given compact DVR. In this way, we have a lot of flexibility in constructing Dedekind domains in over-rings of $\mathbb{Z}[X]$.

I've written out proofs of these statements. As it is much too long to fit here, I'll link to my write-up: Constructing Dedekind domains with prescribed prime factorizations and class groups. Hopefully there's no major errors. I'll also mention that this is an updated and hopefully rather clearer write-up than my initial link (which were very rough notes skipping over many steps).

I think also that my linked proof can be modified to show that you can simultaneously choose any prescribed unit group of the form $\{\pm1\}\times U$ where $U$ is a countable free abelian group.